The question of the first year of junior high school. Let s take a look at it in detail.。。。 Kneel do

1' wheat, sunflower, water chestnut, aquatic weed, bird.

Because a is a negative number, it is preceded by 10 negative signs, which are still negative at this time, and the 3 positive signs are still negative.

The first 11 negative signs are positive at this time, and the 4 positive signs are still positive after simplification.

An even number of negative signs becomes positive, so 10 negative signs, 3 positive signs, the result of simplification is that it itself is negative.

The odd number of negative signs is still a negative sign, so 11 negative signs and 4 positive signs are simplified to -a is a positive number.

Positive signs do not affect the positive or negative values.

a is a negative number.

10 negative signs Because negative negative is positive, then it is a positive sign, and no matter how many positive signs there are, it is also a positive sign, so it is simplified to a negative number.

The 11 minus signs are reduced to positive 1

The first is negative, the second is positive (just look at the negative sign, don't think about the positive sign).

The previous negative number is followed by a positive number.

1: Solution, let the time taken be t, run in the same direction: 350t-250t=400*2 to solve t=8s

Run 350t+250t=400*2 in the opposite direction to solve t=4 3s:22: (1) Set the time taken to t hours, 72t+48t=90 to solve t=3 4 hours.

2) Set the time taken to t hours.

15 minutes = 1 4 hours, 72 * (1 4 + t) + 48 t = 90 solution t = 3 5 hours.

Trapezoidal ABCD, AD BC is known

ed=ec∠edc=∠ecd

eda=∠ecb

Take the DC midpoint M chain to connect ME

Because EDC is an isosceles triangle with three lines in one, EM DC and AD BC are known by the trapezoidal median.

e is the midpoint of ab.

AC=CB ABC is an isosceles triangle and the three-in-one CE is the middle line, so it is also the angular bisector of BCA.

So bce ace= ade

i.e. ade= ace

I'm thinking about this question now, and I don't feel like it's going to be ( . It seems that the auxiliary line can be solved, and it seems that the similarity can be solved, but it is just a little bit, and if you add a condition, it should be simple. There's nothing wrong with the third floor.

Sorry for thinking wrong.

I withdraw my answer.

I know how to answer, but little girl, you just do your homework?

This question** was tested? Dear, did you draw a picture?

og bc, the point o is the intersection of the three angular bisectors of abc, 2=90°- bce=90°-1 2 acb, 1= dab+ dba

1/2(∠bac+∠abc)

1 2 (BAC+ABC+ ACB-ACB) (triangle inner angle and 180°).

90-1/2∠acb=∠2

Because og bc, the point o is the intersection of the three angular bisectors of abc so 2=90°- bce=90°-1 2 acb so 1= dab+ dba

1/2(∠bac+∠abc)

1 2 ( BAC + ABC + ACB - ACB) (triangle inner angle sum) = 90-1 2 ACB = 2